What is the work done in moving a particle around a closed curve?

[bigevil]

Homework Statement

For the field \bold{F} = (y+z) \bold{i} - (x+z) \bold{j} + (x+y) \bold{k} find the work done in moving a particle around the following closed curve:

from the origin to (0,0,2π) on the curve x=1-cos t, y=sin t, z=t; and back to the origin along the z-axis. The answer is 2π. (This question is from Mary L Boas' textbook btw, which is why I have the answer.)

Homework Equations

We get:
dx = sin t dt
dy = cos t dt
dz = dt

x = 1 - cos t
y = sin t
z = t

I did this two ways and didn't get either right. Could someone point out what was wrong with both methods and give me a hint?

First method: I assumed that dx = dy = 0. This gives \int_0^{2\pi} (1-cos t + sin t) dt = 2\pi. But going by this method, going back to origin means that the integral for the reverse action is -2π, so the work is 0. This method is probably wrong because, obviously, x=1 - cos t is not a "static" function.

Second method: Expand everything in full, which means

W = \int F . dr = \int (sin t + t)(sin t) dt - (1 - cos t + t)(cos t) dt + (1 - cos t + sin t) dt = \int_0^{2\pi} 2 + sin t - 2 cos t dt = 4\pi, which is also wrong. For the reverse action, the limits are flipped and I get 0 again.

I think this is the right method, but there's something wrong with the integration here although its supposed to be pretty simple.

 

[tiny-tim]

Hi bigevil!

dx = sin t dt
dy = cos t dt
dz = dt
…
W = \int F . dr = \int (sin t + t)(sin t) dt - (1 - cos t + t)(cos t) dt + (1 - cos t + sin t) dt = \int_0^{2\pi} 2 + sin t - 2 cos t dt

what happened to tsint and tcost?

also, dr² = … ?

 

[bigevil]

Hi bigevil! what happened to tsint and tcost?

also, dr² = … ?

Hi Tim, what do you mean dr^2? There shouldn't be a square in the line integral (F.dr) right?

I think I saw where I got it wrong... I did integration by parts for t cos t and got 0 (between 2π and 0)... then I removed the t cos t / t sin t from the expression. But the integration by parts for t sin t is 2π.

So, 4π - 2π = 2π, but I have another problem, that integral was just one direction. In the other direction the integral is -2π, so the ans. is 0!

 

[tiny-tim]

Hi Tim, what do you mean dr^2? There shouldn't be a square in the line integral (F.dr) right?

You need to know what dr is in terms of t …

I only asked about dr squared because it's easier to calculate!

So, 4π - 2π = 2π, but I have another problem, that integral was just one direction. In the other direction the integral is -2π, so the ans. is 0!

What do you mean, "the other direction"?

That's the negative z-direction, and F.k = x + y.